Prove there are gaps

No.13184690 ViewReplyOriginalReport
if q in Q
and in Q
then q + p in Q: x/y + z/w = (xw + zy) / wy
then q/p in Q.
then (q + p) / 2 in Q.
This says that for every two rational number there is a rational number in between the two.
if A_(1/2)(q, p) = (q + p) / 2 = r then as the recursion of A_(1/2)(q, r) approaches infinity the resulting number is the number directly next to q.
The rational algorithm:
x^2 = 2
if A_(1/2)(1,3)^2 = 2 then x = A
if not then if A^2 > 2 then q = q and p = A
else if A^2 < 2 then q = A and p = p.
under an infinite recursion there will be two numbers containing x and x will be rational and x^2 = 2.
irrational numbers are schizo.